Lời giải:

\[{n_} = \frac{{6,72}}{{22,4}} = 0,3(mol)\]

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

\[{n_{Al}} = \frac{2}{3}{n_} = \frac{2}{3}.0,3 = 0,2(mol)\]

\[ \to {m_{Al}} = 0,2.27 = 5,4(g)\]

m_Cu = 10 – 5,4 = 4,6 (g)

\[\begin{array}{l}\% {m_{Al}} = \frac{{5,4}}.100 = 54\% \\\% {m_{Cu}} = 46\% \end{array}\]

\[\begin{array}{l}{n_{{H_2}S{O_4}}} = 1,5{n_{Al}} = 0,3(mol) \to {m_{{H_2}S{O_4}}} = 29,4(g)\\{m_{{\rm{dd}}{H_2}S{O_4}}} = \frac{{29,4.100}} = 147(g)\end{array}\]